Hi,I am new to pharmacology, I got one problem task ant its all about the dissociation constant of drug (Kd not pKa), So here is the question,If the drug molecular weight is 225.2 grams/mole and the drug dose is 800 grams per day, considering body weight as 60 kg and 70 % water. How to calculate the solubility in grams per liter and in molarity as well as dissociation constant.According to me,1) volume of water present in the 60kg body is 60*70/100 = 42 liters. 2) So if 800 grams is administered concentration present in one liter will be 800/42 = 19.047 mg/l or 0.019 g/l. 3)which i than converted in to moles by dividing 0.019 by its molecular weight that is 225.2 to give 45.8 micromoles.

SO from the above calculation is 0.0019 g/l the solubility and is 45.8 the dissociation constant theoritically.

Or in the second step i have to divide 800 by 1 liters instead of 42.

The solubility of the drug (Acyclovir) in literature comes to 1.3 mg/ml i am really confused can any one please help me?Thanking you in advance.

No, you should dividy it by the whole volume (that is 42 l), the solubility may be whatever, but you are calculating the concentration, not solubility At least, you have lower concentration, than is solubility, what is good.

But I don't think, there is correlation between actuall concentration of medicament and the pKd. I mean, there is, but I don't think, that the level is exactly of pKd, is it?

So as the given solubility of the drug is 1.3 g/l and the calculation shows that on administration of drug = 800 grams, only 0.019 gram is present in one liter, which is less than 1.3 g/l, so what does it indicate.

I mean to say, what happens if the drug concentration is higher than the given solubility and/or lower than given solubility (as in my case)?